According to flightstats.com, American Airfines flights from Dallas to Chicago are on time $80 \%$ of the time. Suppose 10 flights are randomly selected, and the number of on-time flights is recorded.
(a) Explain why this is a binomial experiment.
(b) Determine the values of $n$ and $p$.
(c) Find and interpret the probability that exactly 7 flights are on time.
(d) Find and interpret the probability that fewer than 7 flights are on time.
(e) Find and interpret the probability that at least 7 flights are on time.
(f) Find and interpret the probability that between 5 and 7 flights, inclusive, are on time.
H. Each trial depends on the previous trial.
(b) Using the binomial distribution, determine the values of $n$ and $p$
\[
\begin{array}{l}
n=10 \text { (Type an integer or a decimal. Do not round) } \\
p=0.8 \text { (Type an integer or a decimal. Do not round) }
\end{array}
\]
(c) Using the binomial distribution, the probability that exactly 7 flights are on time is $0.2013^{\text {' }}$
(Round to four decimal places as needed)
Interpret the probability.
In 100 trials of this experiment, it is expected that about $\square$ will result in exactly 7 flights being on time
(Round to the nearest whole number as needed)

Step 1 ：This is a binomial experiment because there are only two possible outcomes (a flight being on time or not), the probability of a flight being on time is constant at 80%, and the flights are independent of each other.

Step 2 ：The values of n and p for the binomial distribution are given in the question. n is the number of trials, which is 10, and p is the probability of success, which is 80% or 0.8.

Step 3 ：To find the probability of exactly 7 flights being on time, we can use the formula for the binomial probability: \(P(X = k) = C(n, k) * (p^k) * ((1 - p)^(n - k))\), where \(P(X = k)\) is the probability of k successes in n trials, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success, and n and k are the number of trials and successes, respectively.

Step 4 ：The probability of exactly 7 flights being on time is approximately 0.2013.

Step 5 ：To interpret the probability, we can multiply the probability by 100 to get the expected number of times out of 100 trials that exactly 7 flights will be on time.

Step 6 ：In 100 trials of this experiment, it is expected that about 20 will result in exactly 7 flights being on time.

Step 7 ：Final Answer: The probability that exactly 7 flights are on time is approximately \(\boxed{0.2013}\). In 100 trials of this experiment, it is expected that about \(\boxed{20}\) will result in exactly 7 flights being on time.